﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 1.1

Question-1 :-  Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225,  (ii)196 and 38220,  (iii)867 and 255

Solution :-
```(i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45.
Therefore, the HCF of 135 and 225 is 45.
```
```(ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196.
```
```(iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops. Since the divisor at this stage is 51.
Therefore, HCF of 867 and 255 is 51.
```

Question-2 :-  Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.

Solution :-
``` Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.Therefore,
a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also,
6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,or 6q + 5
```

Question-3 :-  An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution :-
``` HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
```

Question-4 :-  Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Solution :-
``` Let a be any positive integer and b = 3. Then a = 3q + r for some integer q ≥ 0 And r = 0, 1, 2
because 0 ≤ r < 3 Therefore, a = 3q or 3q + 1 or 3q + 2 Or,
a² = (3q)² or (3q+1)² or (3q+2)²
a² = 9q² or 9q²+6q+1 or 9q²+12q+4
= 3 x 3q² or 3(3q²+2q)+1 or 3(3q²+4q+1)+1
= 3k₁ or 3k₂+ 1 or 3k₃+2
Where k₁, k₂, and k₃ are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

```

Question-5 :-  Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution :-
``` Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,
a³ = (3q)³ = 27a³ = 9(3q³) = 9m
Where m is an integer such that m = 3q³
Case 2: When a = 3q + 1, a³ = (3q +1)³
a³ = 27q³ + 27q² + 9q + 1
a³ = 9(3q³ + 3q² + q) + 1
a³ = 9m + 1
Where m is an integer such that m = (3q³ + 3q² + q)
Case 3: When a = 3q + 2, a³ = (3q +2)³
a³ = 27q³ + 54q² + 36q + 8
a³ = 9(3q³ + 6q² + 4q) + 8
a³ = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
```
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